Question

Does a photon of visible light \((\lambda \approx 400 \text { to } 700 \mathrm{nm})\) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

Short answer

A photon of visible light with a wavelength range of \(400 nm\) to \(700 nm\) has an energy ranging between \(2.846 \times 10^{-19} J\) and \(4.974 \times 10^{-19} J\). These energies are insufficient to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state but can excite an electron from the \(n=2\) to the \(n=6\) energy state.

Step-by-step solution

Calculate the energy of a photon

The energy of a photon can be calculated using the formula: \[E = \dfrac{hc}{\lambda}\] where \(E\) is the energy of the photon, \(h\) is the Planck's constant \((h \approx 6.63 \times 10^{-34} \mathrm{Js})\), \(c\) is the speed of light \((c \approx 3 \times 10^8 \mathrm{m/s})\), and \(\lambda\) is the wavelength of the photon. We are given that the range of wavelengths for visible light is from \(400 nm\) to \(700 nm\). We must calculate the energies corresponding to these wavelengths. First let's convert the wavelengths to meters by multiplying them by \(1\times 10^{-9} m\): \[\lambda_1 = 400\ nm \times(1\times 10^{-9} m/nm) = 4 \times 10^{-7} m\] \[\lambda_2 = 700\ nm \times (1\times 10^{-9} m/nm) = 7 \times 10^{-7} m\] Now, calculate the energy corresponding to \(\lambda_1 = 4 \times 10^{-7} m\): \[E_1 = \dfrac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{4 \times 10^{-7} m} = 4.974 \times 10^{-19} J\] Calculate the energy corresponding to \(\lambda_2 = 7 \times 10^{-7} m\): \[E_2 = \dfrac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{m/s}}{7 \times 10^{-7} m} = 2.846 \times 10^{-19} J\] The energy of the photon varies between \(2.846\times 10^{-19} J\) and \(4.974\times 10^{-19} J\).

Calculate the energy required to excite an electron from \(n=1\) to \(n=5\)

The energy levels of a hydrogen atom can be calculated using the formula: \[E_n = -\dfrac{13.6 \mathrm{eV}}{n^2}\] Thus, the energy required to excite an electron from \(n=1\) to \(n=5\) can be calculated as: \[E_{1 \rightarrow 5} = E_5 - E_1 = \left(\dfrac{-13.6 \mathrm{eV}}{5^2} \right) - \left(\dfrac{-13.6 \mathrm{eV}}{1^2}\right) = 12.09 \text{ eV}\] Now convert this energy to joules (using \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)): \[E_{1 \rightarrow 5} = 12.09 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 1.934 \times 10^{-18} J\]

Calculate the energy required to excite an electron from \(n=2\) to \(n=6\)

Using the same formula as in Step 2, we calculate the energy required to excite an electron from \(n=2\) to \(n=6\): \[E_{2 \rightarrow 6} = E_6 - E_2 = \left(\dfrac{-13.6 \mathrm{eV}}{6^2} \right) - \left(\dfrac{-13.6 \mathrm{eV}}{2^2}\right) = 3.40 \text{ eV}\] Now convert this energy to joules: \[E_{2 \rightarrow 6} = 3.40 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 5.44 \times 10^{-19} J\]

Compare the photon's energy with the required energy for each case

Compare the energy of the photon with the required energy for \(n=1\) to \(n=5\) transition: \[ \begin{cases} 2.846\times 10^{-19} J < 1.934\times 10^{-18} J \\ 4.974\times 10^{-19} J < 1.934\times 10^{-18} J \end{cases} \] Both photon energies are insufficient to excite an electron from the \(n=1\) to the \(n=5\) energy state. Now, compare the energy of the photon with the required energy for \(n=2\) to \(n=6\) transition: \[ \begin{cases} 2.846\times 10^{-19} J < 5.44\times 10^{-19} J \\ 4.974\times 10^{-19} J > 5.44\times 10^{-19} J \end{cases} \] The photon with energy \(4.974\times 10^{-19} J\) has sufficient energy to excite an electron from the \(n=2\) to the \(n=6\) energy state.

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest atom found in the universe, consisting of just one proton and one electron. It is a fundamental building block in chemistry and physics, particularly when studying atomic structures and electron behaviors. In the hydrogen atom, an electron orbits the nucleus at specific energy levels, or shells. These levels are designated by a quantum number, often represented by \(n\).
- A hydrogen atom in its ground state has its electron in the lowest energy level, \(n=1\). - When the electron absorbs energy, it can transition to a higher energy state, moving to an orbit farther from the nucleus.
Understanding the behavior of electrons in a hydrogen atom is essential for grasping quantum mechanics and spectral analysis.

Electron Transition

Electron transitions occur when an electron in an atom moves from one energy level to another. This process can absorb or release energy, depending on the direction of the transition. When an electron absorbs energy, it moves to a higher energy state, often represented by a larger quantum number. Conversely, when it loses energy, it falls to a lower energy state.
- Energy levels in a hydrogen atom are quantized, meaning electrons can't exist between levels but only at specific energies. - For example, moving an electron from \(n=1\) to \(n=5\) requires a specific amount of energy.
Calculating the energy needed for these transitions requires understanding the energy differences between levels. The Balmer series describes the visible spectrum for hydrogen and can be used to understand electron transitions involving visible light.

Visible Light Spectrum

The visible light spectrum consists of the light detectable by the human eye, spanning wavelengths approximately between 400 nm and 700 nm. This spectrum forms a small part of the electromagnetic spectrum and includes all the colors seen in a rainbow from violet to red.
- Photons of visible light carry different amounts of energy depending on their wavelength. Shorter wavelengths, like violet, carry more energy than longer wavelengths like red. - Specific transitions in the hydrogen atom correspond to wavelengths in this spectrum, which produce visible colors.
Examining the energy associated with these photons is crucial, particularly when testing their ability to induce electron transitions in atoms. Knowing the energy range of visible light helps in understanding which photon energies can cause these transitions.

Planck's Constant

Planck's constant, denoted as \(h\), is a fundamental constant in quantum mechanics with a value of approximately \(6.63 \times 10^{-34}\;\text{Js}\). It plays a critical role in the quantification of energy at the atomic level.
- Planck's constant is used in the equation \(E = \frac{hc}{\lambda}\), where \(E\) is the energy of a photon, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon. - This equation reveals how energy is inversely proportional to the wavelength: as the wavelength decreases, energy increases.
Understanding Planck's constant is essential for calculations involving photon energy, electron transitions, and the analysis of spectra. Its introduction marked a huge leap in understanding quantum phenomena and remains pivotal in modern physics.